Question: A rectangular loop has a sliding connector PQ of length $\ell$ and resistance $R \Omega$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $\mathrm{I}_{1}, \mathrm{I}_{2}$ and I are :-
$\mathrm{I}_{1}=\mathrm{I}_{2}=\frac{\mathrm{B} \ell \mathrm{v}}{6 \mathrm{R}}, \mathrm{I}=\frac{\mathrm{B} \ell \mathrm{v}}{3 \mathrm{R}}$
$\mathrm{I}_{1}=-\mathrm{I}_{2}=\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}}, \mathrm{I}=\frac{2 \mathrm{~B} \ell \mathrm{v}}{\mathrm{R}}$
$\mathrm{I}_{1}=\mathrm{I}_{2}=\frac{\mathrm{B} \ell \mathrm{V}}{3 \mathrm{R}}, \mathrm{I}=\frac{2 \mathrm{~B} \ell \mathrm{V}}{3 \mathrm{R}}$
$\mathrm{I}_{1}=\mathrm{I}_{2}=\mathrm{I}=\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}}$
Correct Option: , 3
Solution:
